Math Insight

Integration substitutions


The chain rule can also be ‘run backward’, and is called change of variables or substitution or sometimes u-substitution. Some examples of what happens are straightforward, but others are less obvious. It is at this point that the capacity to recognize derivatives from past experience becomes very helpful.

Example Since (by the chain rule) $${d\over dx} e^{\sin x}=\cos x\;e^{\sin x}$$ then we can anticipate that $$\int \cos x\;e^{\sin x}\; dx= e^{\sin x}+C$$

Example Since (by the chain rule) $${d\over dx} \sqrt{x^5+3x}={1\over 2}(x^5+3x)^{-1/2}\cdot(5x^4+3)$$ then we can anticipate that $$\int {1\over 2}(5x^4+3)(x^5+3x)^{-1/2}\;dx=\sqrt{x^5+3x}+C$$

Very often it happens that things are off by a constant. This should not deter a person from recognizing the possibilities. For example: since, by the chain rule, $${d\over dx}\sqrt{5+e^x}={1\over 2}(5+e^x)^{-1/2}\cdot e^x$$ then $$\int e^x\,(5+e^x)^{-1/2}\;dx=2\int {1\over 2}e^x(5+e^x)^{-1/2}\;dx= 2\,\sqrt{5+e^x}+C$$ Notice how for ‘bookkeeping purposes’ we put the ${1\over 2}$ into the integral (to make the constants right there) and put a compensating $2$ outside.

Example: Since (by the chain rule) $${d\over dx}\sin^7(3x+1)=7\cdot \sin^6(3x+1)\cdot \cos (3x+1)\cdot 3$$ then we have \begin{multline*}\int \cos(3x+1)\,\sin^6(3x+1)\;dx\\ ={1\over 21}\int 7\cdot 3\cdot \cos(3x+1) \sin^6(3x+1)\; dx= {1\over 21} \sin^7(3x+1)+C \end{multline*}


  1. $\int \cos x\,\sin x\,dx=?$
  2. $\int 2x \;e^{x^2}\;dx=?$
  3. $\int 6x^5\;e^{x^6}\; dx=?$
  4. $\int { \cos x \over \sin x }\; dx=?$
  5. $\int \cos x e^{\sin x}\; dx=?$
  6. $\int { 1 \over 2\sqrt{x }}e^{\sqrt{x}}\; dx=?$
  7. $\int \cos x \sin^5 x\; dx=?$
  8. $\int \sec^2 x \tan^7 x\; dx=?$
  9. $\int (3\cos x+ x)\,e^{6\sin x+x^2}\,dx=?$
  10. $\int e^x\sqrt{e^x+1}\,dx=?$